# need help with limit, just a refresher

• Sep 22nd 2009, 10:58 AM
ffxwolf
need help with limit, just a refresher
Prove that the stated limit is correct by finding a number d (delta) in terms of E (is an element of) such that
|f(x) - L| < E whenever |x - c|< d:

lim (4x - 1) = 11
x -> 3
• Sep 22nd 2009, 11:28 AM
redsoxfan325
Quote:

Originally Posted by ffxwolf
Prove that the stated limit is correct by finding a number d (delta) in terms of E (is an element of) such that
|f(x) - L| < E whenever |x - c|< d:

lim (4x - 1) = 11
x -> 3

You want to find $\displaystyle \delta$ such that $\displaystyle |x-3|<\delta \implies |(4x-1)-11|<\epsilon$.

Choose $\displaystyle \delta=\frac{\epsilon}{4}$.

Thus when $\displaystyle |x-3|<\delta$, we have that $\displaystyle |(4x-1)-11|=|4x-12|=4|x-3|<4\delta=\frac{4\epsilon}{4}=\epsilon$
• Sep 22nd 2009, 11:43 AM
ffxwolf
Thank you for your response, I just don't understand why delta = epsilon/4
• Sep 22nd 2009, 11:53 AM
redsoxfan325
Quote:

Originally Posted by ffxwolf
Thank you for your response, I just don't understand why delta = epsilon/4

If you don't know what $\displaystyle \delta$ is, you can solve for it.

Whatever $\displaystyle \delta$ is you know that $\displaystyle |x-3|<\delta$.

So, work with what you have. You ultimately want $\displaystyle |(4x-1)-11|<\epsilon$.

So, similar to before,

$\displaystyle |(4x-1)-11|=|4x-12|=4|x-3|<4\delta$.

Now you want $\displaystyle 4\delta$ to be equal to $\displaystyle \epsilon$, so you just solve for it.

$\displaystyle 4\delta=\epsilon\implies \delta=\frac{\epsilon}{4}$
• Sep 22nd 2009, 11:58 AM
ffxwolf
Okay, I see it now. Thanks (Nod)